Discussion
A device has 20 address lines and an 8-bit data bus. Its directly addressable memory is
Answer: C
Correct answer: 1 MB
\[ 2^{20} = 1{,}048{,}576 \] addressable locations, each holding one byte on an 8-bit data bus, so 1 MB. The data bus width sets the size of each location, and the number of address lines sets how many there are - multiplying the two, as the 8 MB option does, confuses bits with bytes.
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